Is the Hausdorffness a necessary condition for a manifold to be smooth? If not, why we include Hausdorff condition in the definition of differentiable manifold?
NO, we do not need Manifold to be Hausdorff space. There are non Hausdorff smooth manifolds. For example take RX{0}URX{1}. Consider the subspace topology. Now identify (x,0) and (x,1) for x
@Subrata Bhowmik
As you cited the example, any neighbourhood N of (0,0) is homeomorphic to some open interval in R. Now, consider any such N and remove the point (0,0) so that we will get 3 connected components while we will get 2 connected components if we remove the image point of (0,0) from the image of N. So, the number of connected components will not be preserved. Isn't it a absurd behaviour for a smooth manifold?
See the exercises 2, page 64, in " An introduction to differentiable manifolds and Riemannian geometry" by W, M. Boothby.
##Dr. Subrata, thank you so much for you r explanation, but may you kindly comment on the Dr. Zafar's view.
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